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本帖最后由 hbghlyj 于 2023-4-15 13:25 编辑 Fubini's Theorem: Let \(f:ℝ^2→ℝ\) be integrable. Then for almost all \(y∈ℝ\), \(x↦f (x, y)\) is integrable in \(x\) and defining \(F (y)=\int_ℝf (x, y) \mathrm{d} x\), \(F\) is integrable and\[\int_{ℝ^2} f (x, y) \mathrm{d} (x, y)=\int_ℝF (y) \mathrm{d} y=\int_ℝ\left(\int_ℝf (x, y) \mathrm{d} x\right) \mathrm{d} y\]
1) By assumption $y\mapsto f(y)$ is integrable, $x\mapsto xf(x)$ is integrable.
$$\int_{-\infty}^{\infty} f(x) d x=1, F(x)=\int_{-\infty}^x f(y) d y\implies 1-F(x)=\int_x^{\infty} f(y) d y$$
Applying Fubini's Theorem
$$\int_0^{\infty}(1-F(x)) d x=\int_0^{\infty}\int_x^{\infty} f(y) dydx=\int_0^{\infty}\int_0^yf(y) dxdy=\int_0^{\infty}yf(y) dy$$
and
$$\int_{-\infty}^0 F(x) d x=\int_{-\infty}^0\int_{-\infty}^x f(y) d ydx=\int_{-\infty}^0\int_y^0 f(y) d xdy=\int_{-\infty}^0(-y)f(y) dy$$
2) We have $1-G(y)-G(-y)=\int_{\{g(x)>y\}} f(x) d x-\int_{\{g(x)\le -y\}} f(x) d x$
\begin{align*}\int_0^{\infty}(1-G(y)-G(-y)) d y&=\int_{0}^{\infty}\int_{\{g(x)>y\}} f(x) d xdy-\int_{0}^{\infty}\int_{\{-g(x)\ge y\}} f(x) d xdy\\
&=\int_{\{g(x)>0\}}\int_0^{g(x)}f(x) dydx-\int_{\{g(x)\le 0\}}\int^{-g(x)}_0f(x) dydx\\
&=\int_{\{g(x)>0\}}f(x)g(x)dx+\int_{\{g(x)\le 0\}}f(x)g(x)dx\\
&=\int_{-\infty}^\infty f(x)g(x)dx
\end{align*} |
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